How to remove a node from a singly-linked list when only given the pointer to the node

Question

jobseeker · Accepted Answer

thisNode.data = thisNode.next.data
temporaryNode = thisNode.next.next
delete thisNode.next
thisNode.next = temporaryNode

anon · Answer

Milly: jobseeker is right in that you don't delete the "node to be deleted". You delete the node that FOLLOWS the "node to be deleted", only after swapping the values of the two.

Milly · Answer

The post by "jobseeker" assumes that we know the node BEFORE the node to be deleted. This assumption is not consistent with the problem definition, but makes it solvable.

Anónimo · Answer

Milly, jobseeker's assumption is reasonable, we should not delete the node before removing it from the list (if it was created by the list)

David Dhas · Answer

node ->val=node->next->val;
     node ->next=node->next->next;

Anónimo · Answer

node = node.next;

ashu · Answer

node.val = node.next.val;
  node.next = node.next.next;

Anónimo · Answer

You have to use some pointer/memory trickery

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